logic design dr. yosry a. azzam. 2 boolean algebra & logic gates chapter 2

Logic DesignLogic Design

Dr. Yosry A. AzzamDr. Yosry A. Azzam

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Boolean Algebra & Logic Gates

Chapter 2

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Agenda Basic definitions Proprieties of Boolean

Algebra Boolean Functions Canonical and standard

Forms

Reading• Mano: Ch 2

Objectives Understanding the

canonical forms Maxterms and minterms Simplifying Boolean

expression and functions

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Boolean AlgebraThe mathematical system that operate with

binary values (George Boole (1854) )Is the algebraic structure defined on a set of

elements with two binary operators + (OR) and . (AND) provided that the following Huntington postulates are satisfied:

Reading :Mano Chapter 2

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Huntington Postulates1. (a) closures w.r.t operator + (b) closures w.r.t operator .2. (a) An identity element w.r.t + (0): x + 0 = 0 + x = x (b) An identity element w.r.t . (1) : x . 1 = 1 . x = x3. (a) Commutative w.r.t +: x + y = y + x (b) Commutative w.r.t . : x . y= y . x4. (a) . is distributive over +: x . (y + z) = ( x . y) + ( x . z) (b) + is distributive over .: x + (y . z) = (x + y) . (x + z)5. The complement of x is such that (a) x + x’ =1 and (b) x . x’ =06. There exist at least 2 elements x , y such that x ≠ y

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Two-Valued Boolean Algebra Is defined on a set of 2 elements, B = {0,1} with

rules for the two binary operators + and . as shown in the following tables:

These rules are exactly the same as the AND, OR, and Not operations.

X Y X.Y

0 0 0

0 1 0

1 0 0

1 1 1

X Y X+Y

0 0 0

0 1 1

1 0 1

1 1 1

X X’

0 1

1 0

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Huntington Postulates on the Two-Valued Set and the two Binary Operators1. closures is obvious from the tables as the result of each operation is

either 0 or 12. (a) An identity element w.r.t + (0): 0 + 0 = 0 , 0+1 = 1+0 =1 (b) An identity element w.r.t . (1) : 1 . 1 = 1 , 1 . 0 = 0 . 1 =03. The Commutative laws are obvious4. (a) the distributive law : x . (y + z) = ( x . y) + ( x . z) can be verified

by the truth table of all possible values of x, y and z (b) the distributive law: x + (y . z) = (x + y) . (x + z) can also be

verified by the truth table of all possible values of x, y and z5. The complement of x (a) x+x’ =1: 0+0’=0+1=1 and 1+1’ =1+0 =1

(b) x . x’ =0: 0.0’=0.1=0 and 1.1’ =1.0 =06. The two valued Boolean algebra has two distinct elements 1 and 0

with 1 ≠0

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Truth Tables Verification

x y z y + z x . ( y + z )

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 1 0

1 0 0 0 0

1 0 1 1 1

1 1 0 1 1

1 1 1 1 1

x y z x . y x . z (x . y)+( x . z )

0 0 0 0 0 0

0 0 1 0 0 0

0 1 0 0 0 0

0 1 1 0 0 0

1 0 0 0 0 0

1 0 1 0 1 1

1 1 0 1 0 1

1 1 1 1 1 1

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Property of Boolean Algebra 1Closure

Obtaining a unique elements (which are the members of Boolean set)

Associative Law (X*Y)*Z = X*(Y*Z) (X+Y)+Z = X+(Y+Z)

Commutative Law X*Y = Y*X X+Y = Y+X

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Property of Boolean Algebra 2

Identity Element 1.X=X.1=X 0+X=X+0=X

Inverse X. X’ =0 X+ X’ =1

Distributive LawX+(Y.Z)=(X+Y).(X+Z)

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Basic Theorems

Duality Huntington (1904) postulate that of an

algebraic expression, we can simply interchange OR and AND operator and replace 1 by 0 and 0 by 1

[We will discuss this more in future sessions when you enter the realm of digital design]

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Postulates and Theorems of Boolean Algebra

Postulate 2 (a) x + 0=x (b) x .1=x

Postulate 5 (a) x + x’=1 (b) x . x’=0

Theorem 1 (a) x + x=x (b) x . x=x

Theorem 2 (a) x+1=1 (b) x . 0=0

Theorem 3 Involution (x’)’ = x

Postulate 3 Commutative (a) x +y=y +x (b) x y=y x

Theorem 4 Associative (a) x +(y +z)=(x +y) +z (b) x (y z)=(x y) z

Postulate 4 Distributive (a) x (y +z)= x y+ x z (b) x +y z=(x +y) (x +z)

Theorem 5 DeMorgan (a) (x +y)’=x’ y’ (b) (x y)’=x’ + y’

Theorem 6 Absorption (a) x + x y= x (b) X (x +y)=x

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Proof of some basic theorems

Theorem 1(a): x+x=x.Proof:x + x = (x+x).1 (postulate 2b)=

= (x + x) . (x + x') (postulate 5a

= x + xx' = x (postulate 4b)Theorem 1(b): x . x=xProof:x . x = xx +0 = x x + x x' = x . (x+x')= x . 1 = x

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Proof of some basic theorems

Theorem 2(a): ): x +1=1.Proof:x+1= 1 . (x+1) = (x+x')(x+1) = x+x' . 1 = x+x' =1Theorem 2(b): x .0=0.Proof: by duality x.0=0

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Proof of some basic theorems

Theorem 6(a): x+xy=x.Proof:x +x y=x.1+xy=x(1+y)=x (y+1)=x.1=x

Theorem 6(b): x(x+y)=x.Proof: By duality

x y x y x + x y

0 0 0 0

0 1 0 0

1 0 0 1

1 1 1 1

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DeMorgan’s Theorem

x y x + y (x + y)’

0 0 0 1

0 1 1 0

1 0 1 0

1 1 1 0

x’ y’ x’ y’

1 1 1

1 0 0

0 1 0

0 0 0

(a) (x + y)’=x’ y’ (b) (x y)’=x’+ y’

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Operator precedence

ParenthesisNOTANDOR

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Boolean Functions 1

F1=x+y’.z

F2=x.y.z’

X Y Z F1 F2

0 0 0 0 0

0 0 1 1 0

0 1 0 0 0

0 1 1 0 0

1 0 0 1 0

1 0 1 1 0

1 1 0 1 1

1 1 1 1 0

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Boolean Functions 2

HW : give the truth table of Functions F1 and F2

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Algebraic Manipulation Simplify the following Boolean function to a minimum number of

literals:(1) F (x , y) = x (x' + y)(2) F (x , y) = x + x' y(3) F (x , y) = (x + y) (x + y')(4) F ( x , y , z) = x y + x' z + y z(5) F( x, y, z) = (x + y) (x' + z) (y + z) Solution:(1) F (x , y) = x x' + x y = 0 + x y = x y(2) F (x , y) = (x + x') ( x + y) = 1.(x + y) = x + y(3) F (x , y) = x + y y' = x + 0 = x

or= x + x y' + y x + y y' = x (1 + y' + y)= x . 1 = x

(4) F (x , y , z) = x y + x' z + y z (x + x' ) = x y + x' z + x y z + x' y z = x y (1 + z) + x' z (1 + y) = x y + x' z(5) (x + y) (x' + z) (y + z)=(x + y) (x' + z) by duality from 4

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Complement of a FunctionDe Morgan

(X+Y)’ = X’.Y’(A+B+C)’

= (A+X)’ with X=B+C = A’X’ (De Morgan) = A’.(B+C)’ = A’.(B’C’) (De Morgan) = A’.B’.C’ (Associative)

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Complement of a FunctionExample:Find the complement of the functions: (1) F=A+B+C.(2) F1=x' y z' + x' y' z(3) F2= x (y' z' +y z)Solution:(1) F'=(A+B+C)'=A'B'C '(2)F1'=(x' y z' + x' y' z)' = (x' y z')' (x' y' z)' = (x + y' + z) (x + y + z')(3) F2'=x' + (y + z) (y' + z')

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Canonical Forms

Binary Variable can either be: Normal Form (x) Complement Form (x’) For 3 binary variables (x, y, z) there are 8

possibility of response for AND operation (minterms) and 8 possibility of OR operation (maxterms)

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Minterms #1“ Minterms” or “Standard Product”2 Binary Variable (X & Y) will form 2n=2

Minterms X’Y’, X’Y, XY’, XY AND Terms (“product”)

Any Boolean function can be expressed as a sum of Minterms

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Minterms and Maxterms for three binary variables:

Minterms Maxterm

x y Z Term Designation Term Designation

0 0 0 x'y'z' m0 x+y+z M0

0 0 1 x'y'z m1 x+y+z' M1

0 1 0 x'yz' m2 x+y'+z M2

0 1 1 x'yz m3 x+y'+z' M3

1 0 0 xy'z' m4 x'+y+z M4

1 0 1 xy'z m5 x'+y+z' M5

1 1 0 xyz' m6 x'+y'+z M6

1 1 1 xyz m7 x'+y'+z' M7

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Function of three variablesx y Z Function f1 Function f2

0 0 0 0 0

0 0 1 1 0

0 1 0 0 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

F1=x’ y’ z + x y’ z’ + x y z = m1+m4+m7F2=x’ y z + x y’ z + x y z’ + x y z = m3 + m5 + m6 + m7

F’1= x’ y’ z’ + x’ y z’ + x’ y z + x y’ z + x y z’ The complement of F’1=(x+y+z)(x+y’+z )(x+y’+z’)( x’+y+z’) (x’ +y’+z) =F1= M0.M2.M3.M5.M6

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Minterms #2Example 2-4 (page 46)

F = A + B’C F = A (B + B’) + B’C F = AB + AB’ + B’C F = AB (C + C’) + AB’ (C + C’) + B’C F = ABC + ABC’ + AB’C + AB’C’ + B’C F = ABC + ABC’ + AB’C + AB’C’ + (A + A’) B’C F = ABC + ABC’ + AB’C + AB’C’ + AB’C + A’B’C F = ABC + ABC’ + AB’C + AB’C’ + A’B’C F = m7 + m6 + m5 + m4 + m1

(Table 2-5, page 47) F(A, B, C) = ∑(1, 4, 5, 6, 7)

Sum of Minterms

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Truth Table for F=A+B’C

A B C F

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 0

1 0 0 1

1 0 1 1

1 1 0 1

1 1 1 1

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Maxterms“ Maxterms” or “Standard Sums”2 Binary Variables (X & Y) will form 2n=2

Maxterms X’+Y’, X’+Y, X+Y’, X+Y OR Terms (“sum”)

Any Boolean function can be expressed as a product of Maxterms F2=(x+y+z)(x+y+z’)(x+y’+z)(x’+y+z) F2=M0 . M1 . M2 . M4=Π(0,1,2,4)

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Minterms & Maxterms #1

Minterms X’Y’, X’Y, XY’, XY

Maxterms X’+Y’, X’+Y, X+Y’, X+Y

Each Maxterms is the complement of its corresponding Minterms & vice versa Remember De Morgan? Take a look at the next slide

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Minterms & Maxterms #2

Minterms Maxtermsx y z Term Desig. Term Desig.

0 0 0 x’ y’ z’ m0 x + y + z M0

0 0 1 x’ y’ z m1 x + y + z’ M1

0 1 0 x’ y z’ m2 x + y’ + z M2

0 1 1 x’ y z m3 x + y’ + z’ M3

1 0 0 x y’ z’ m4 x’ + y + z M4

1 0 1 x y’ z m5 x’ + y + z’ M5

1 1 0 x y z’ m6 x’ + y’ + z M6

1 1 1 x y z m7 x’ + y’ + z’ M7

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Conversion between canonical formsX Y Z F1

0 0 0 0

0 0 1 1

0 1 0 0

0 1 1 1

1 0 0 0

1 0 1 0

1 1 0 1

1 1 1 1

F=xy + x’z

The function F expressed in sum of minterms is: F(x,y,z)= ∑(1,3,6,7)

The missing terms are 0,2,4,5

Then, The function F expressed in product of maxterm is:

F(x,y,z)= Π(0,2,4,5)

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Standard Forms

Doesn’t have to consists of all variables Sum of Products : F1 = y’ + xy + x’yz’ Products of Sum : F2 = x (y’ + z) (x’ + y + z’)

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Implementation with two and three levels

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Truth table for 16 functions of 2 binary variables

X Y F0 F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 F12 F13 F14 F15

0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

0 1 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

1 0 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

1 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

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Other Logic OperationsBoolean Operator Operator Symbol Name Comments

F0 = 0 Null Binary Constant 0

F1 = xy x.y AND x and y

F2 = xy’ x/y Inhibition x but not y

F3 = x Transfer x

F4 = x’y y/x Inhibition y but not x

F5 = y Transfer y

F6 = xy’ + x’y xy Exclusive OR x or y but not both

F7 = x + y x+y OR x or y

F8 = (x + y)’ xy NOR Not OR

F9 = xy + x’y’ (xy)’ Equivalence x equals y

F10 = y’ y’ Complement Not y

F11 = x + y’ X y Implication If y then x

F12 = x’ x’ Complement Not x

F13 = x’ + y X y Implication If x then y

F14 = (xy)’ xy NAND Not AND

F15 = 1 Identity Binary Constant 1

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Common Logic Gates

AND OR Inverter Buffer NAND NOR XOR XNOR

Figure 2-5 (page 54)

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Positive and negative logic

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Positive and negative logic (Contd.)

H

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Integrated Circuits (ICs)Levels of Integration1- Small-Scale Integration (SSI)

Number of gates are < 10

2- Medium-Scale Integration (MSI) Number of gates are from 10 to 1000

3- Large-Scale Integration (LSI) contains thousands of gates in a single package

4- Very Large-Scale Integration (VLSI) contains hundred of thousands of gates in a single package

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Digital Logic familiesDigital ICs are classified not only by their

complexity or logical operations but also by the specific circuit technology to which they belong which is referred to as digital logic family.

The basic circuits in each technology is a NAND, NOR, or inverter gate.

The electronic components employed in the construction of the basic circuit are usually used to name the technology.

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Digital Logic families (Contd.)The most popular logic families are:1- TTL Transistor-Transistor Logic

is being in operation for a long time and is a standard

2- ECL Emitter-Coupled Logichas an advantages in systems requiring high speed operation

3- MOS Metal-oxide Semiconductorsuitable for circuits that need high component density

4- CMOS Complementary Metal-oxide Semiconductorpreferable in systems requiring low power consumption as in VLSI

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Digital Logic families parametersThese are the parameters that are evaluated and

compared for different families:1- Fan-out: the number of standard loads that the output of a gate

can drive.

2- Fan-in: the number of inputs available in the gate.

3- Power dissipation: power consumed by the gate that must be available from the power supply.

4- Propagation delay: average transition delay time for the signal to propagate from input to output.

5- Noise margin: is the maximum external noise voltage added to an input signal that does not cause an undesirable change in the circuit output

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Computer-Aided Design (CAD)

Software programs that support computer-based representation and aid in the development of digital hardware by automating the design process.

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Exercises

HW :

Problems 2-1 until 2-23 Mano